a geometry solutions v02

8/18/2019 A Geometry Solutions v02

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SAT

GEOMETRY SOLUTIONSI. Two-Sided Figures

1. (D) 72 The degree measure around any point is

360º, just like a circle. Since there are 5 x’s here, 5 x

= 360º. Therefore x = 72º.

2. (A) 65 The angle that is opposite the 50º angle

is also 50º. That angle and the two angles labeled y

will add up to 180º. Therefore,

2 y + 50º = 180º

2 y = 130º

y = 65º

3. (C) 70 Since the lines are parallel, the angle

adjacent to the angle marked 2 x – 30 will equal x.

Since the degree measure of a line is 180º, an

equation can be formed.

x + (2 x – 30) = 180º

3 x – 30 = 180º

3 x = 210º

x = 70º

4. (B) 5 Since lines 1 and 2 are parallel, and x is

95º, the labeled angle must be 95º (See the figure

below.) Therefore, the labeled supplementary angle

must be 85º.

Since lines 3 and 4 are perpendicular, one of theother angles of the triangle must be a right angle.

All of the angles in a triangle must add up to 180º.

Therefore, the angle y can be found by:

y + 90º + 85º = 180º

y + 175º = 180º

y = 5º

5. (B) 80 Since l and m are parallel and w and b

are alternate interior angles, then w = b = 80º. The

angles x, w, and a must add up to 180º, because the

angles all comprise one side of a line. Since a = 20º

and w = 80º, solve for x.

x + 20º + 80º = 180º

x + 100º = 180º

x = 80º

Since x and y are corresponding angles, then x = y

= 80º. Now that the values for x, y and z are known,

find x + y – z.

x + y – z = 80 + 80 – 80 = 80

95º

85º

1

2

3

4

y°

x°

SECTION AA: Solutions For SAT Geometry

AA-1 Copyright www.TestMasters.com. All rights reserved. AA-1


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6. (C) Six Draw 4 lines following the given rules.

Here is one example with the points of intersection

circled.

If drawn correctly according to the given rules there

will always be six points of intersection.

II. The 4 Types of Triangles

1. (D) 105 Remember that when lines intersect,

the angles opposite each other are equal. Therefore,

the angle opposite the 45º angle will also be 45º.Also remember that the sum of the angles in a

triangle is 180º. Therefore,

30º + 45º + xº = 180º

75º + xº = 180º

xº = 105º

2. (B) 105 If BAC = 80º, then ABC must equal

70º because the sum of the angles of any triangle is

180º. Since xº =1

2 BAC and yº =

1

2 ABC , xº = 40

and yº = 35. Also note that xº + yº + z º = 180.

Substitute in values for x and y and then solve for z .

40º + 35º + z º = 180º

z º = 105º

3. (A) 50 Since AB = BC , the triangle is isosceles.

Therefore, BAC = ACB = xº. Since the sum ofthe angles of a triangle equal 180º, make an

equation.

xº + xº + 80º = 180º

2 xº + 80º = 180º

2 xº = 100º

xº = 50º

III. Area of a Triangle

1. (D) 36 The area of a triangle is equal to1

2bh.

Use this formula to solve for height, given the area

of 60 and the base of 10.

60 =1

2(10)h

60 = 5h

12 = h

Since the height is always perpendicular to the base,it creates a right triangle as shown below. Also, note

that the base of the right triangle will be 5, because

the height bisects the base of the isosceles triangle.

Use Pythagorean Theorem to solve for x.

52 + 122 = x2

25 + 144 = x2

169 = x2

13 = x

The perimeter equals x + x + 10 or 13 + 13 + 10 =

36.

2. (A) AOX Since all the triangles have the

same base (OX ), the determining factor is the height.

Since AOX has the least height, it also has the

least area.

IV. The Right Triangle

1. (C) 30 + 12 2 The figure is comprised of two

right triangles. The height of the figure can be found

by using Pythagorean Theorem with the base and

hypotenuse of the left-side triangle. The height ofthe figure is labeled as h.

5

12 x




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52 + h2 = 132

25 + h2 = 169

h2 = 144

h = 12

The right-side triangle has one 90º angle and one 45º

angle. Therefore, the remaining angle must be 45ºand the triangle is a 45º-45º-90º triangle. Since it is a

45º-45º-90º triangle, the remaining two sides of the

triangle are 12 and 12 2 . See the figure below for

all labeled sides.

The perimeter will 5 + 12 + 13 + 12 2 =

30 + 12 2 .

2. (D) 30 3

The area of a triangle is1

2bh. First,

find the height of the triangle. Since the height is

always perpendicular to the base, labeling the height

creates a right triangle. One of the angles is 60º and

the other is 90º, so the remaining angle is 30º.

Therefore, the triangle is a 30º-60º-90º triangle asshown below.

The height of this 30º-60º-90º triangle must be 4 3

because is corresponds to the 60º angle. The area of

the triangle can be found using the formula.

A =1

2bh

A =1

2(15)( 4 3 )

A = (15)( 2 3 )

A = 30 3

V. Similar Triangles

1. (A) 1 Because the two triangles share two

angles (90º and xº), they are similar. Because BC =

CD, the two triangles are not only similar, they are

identical because those sides correspond to oneanother. Since ABC is a right triangle, use the

Pythagorean Theorem to solve for BC .

32 + BC 2 = 52

9 + BC 2 = 25 BC 2 = 16

BC = 4

Since the triangles are identical, then AC = CE = 3.

The length of BE will be equal to the length of BC

minus the length of CE .

BE = BC – CE

BE = 4 – 3 = 1

2. (D) 12.5 Set up a proportion of height raised to

horizontal distance of the corresponding sides. Let x

be the height.

1

8 100

x

100 = 8 x

12.5 = x

3. (A) 5 Since AB = 1 and CD = 3, the two

triangles have a ratio of 1 : 3. The line segments x

and 3 x can then be labeled, since the length of the

larger segment must be 3 times the length of the

smaller. Also, label the hypotenuse of the smaller

triangle as c. See the figure below.

12

12

30º

A B

C D

x

3 x

c




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Since x + 3 x is 4 x and the sum of the line segments

is CB which is 3,

4 x = 3

x =

3

4

Since x =3

4, the length of the hypotenuse of the

smaller triangle can be found with the Pythagorean

Theorem.

c2 =

23

4

+ 12

c2 =9

16+ 1

c2 = 2516

c =5

4

Since the hypotenuse of the larger triangle must be 3

times as long, the length is

5

4× 3 =

15

4

The total length of AD is the sum of the hypotenuses

of both triangles.

5

4 +

15

4 =

20

4= 5

ALTERNATE SOLUTION

A right triangle can be created by drawing the lines

shown below.

This right triangle will have a height equal to CB,

which is 3. The base of the triangle will be equal to

AB + CD, which is 1 + 3 = 4. Using the

Pythagorean Theorem, the length of the hypotenuse

AD can be found.

AD2 = 32 + 42

AD2 = 9 +16

AD2 = 25

AD = 5

VI. Three More Rules

1. (A) 21 and 29 The third, unknown side of this

triangle must be greater than the difference of the

two known sides. The third side must be less than

the sum of the two known sides. Therefore, where x

is the third side,

10 – 5 < x < 10 + 5

5 < x < 15

Since the perimeter is an integer (positive or

negative whole number) x must be a whole number.

Since 5 < x < 15, the largest integer value for x is14. The smallest integer value for x is 6.

If x is as large as possible, the perimeter of the

triangle will be 5 + 10 + 14 = 29. If x is as small as

possible, the perimeter of the triangle will be 5 + 10

+ 6 = 21.

VII. Four-sided Figures

1. (D)

2

2 y

The area of a rectangle equals lw and in

this case, l = x and w = y. Therefore the area is equal

to xy. Since x is half of y, or as an equation x =2

y,

the area is equal to

A B

C D




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2

y(y) =

2

2

y

2. (C) 540º The indicated angles are equal to the

interior angles of the pentagon because they are

corresponding angles. The sum of the angles for any

polygon is equal to (180º)(n – 2) where n is the

number of sides. Since n = 5 in this case, the total

number of degrees equals (180º)(3) or 540º.

3. (B) 8 The perimeter of any rectangle equals 2l+ 2w. In this case, l = 3 x and w = 2 x. Plug these

values into the formula and set it equal to 80.

2l + 2w = 80

2(3 x) + 2(2 x) = 806 x + 4 x = 80

10 x = 80

x = 8

4. (C)18 Divide the shape into 3 rectangularregions as shown below.

Area A is equal to 3 × 3 = 9. Area C is equal to 4 ×2 = 8. Next, find the dimensions of area B. The

height must be equal to 3 – 2 = 1. The base must be

equal to 8 – (4 + 3) = 8 – 7 = 1. Therefore area B is

equal to 1 × 1 = 1. The total area of the figure is

equal to A + B + C = 9 + 8 + 1 = 18.

5. (C) 320 To find the number of squares in the

shaded area, divide the shaded area by the area of

each square.

80 100 8,000

5 5 25

= 320

VIII. Circles

1. (A) 9 Find the area of each circle using the

formula “ Area = r 2”.

r = 6 : (6)2 = 36

r = 2 : (2)2 = 4

Then, divide the larger area by the smaller area.

36

4

= 9

2. (A)9

2

The region that is described is a

semicircle (half of a circle) with a radius equal to the

length of AB, 3. See the figure below.

The area of a semicircle is equal to1

2the area of a

circle, which is “r 2”.

2 2(3) 9

2 2 2

r

3. (D) 112.5º Since the indicated angle is 5 marks

wide, and there are a total of 16 marks, the angle

must be5

16

of the degree measure of a circle. A

circle has a degree measure of 360º. To find the

degree measure of this angle, multiply5

16 by

360º.

5

16× 360º = 112.5º

4.(C)3

The length of an arc is equal to the

circumference multiplied by the degree measure of

the arc divided by 360º. The circumference of acircle equals “2r ”. Solve using the given values.

30

360

× 2(2) =

1

12× 4 =

4

12

=

3

5. (B) 4

The polygon is equilateral, meaning all

the sides are equal. The vertices of the polygon are

A B A

180º




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equally spaced apart. Therefore, the vertices divide

the circle in to equal segments. Here, the circle is

divided into 5 equal segments, and the arc PQR is 2

of those segments. The arc is then2

5 of the entire

circumference.

The arc length can then be solved by multiplying2

5

by the circumference. Remember that circumference

can also be written as “d .”

2

5× d =

2

5× 10 =

20

5

= 4

6. (D) 6r 2 The length and width of the rectangle

can be determined by using the radii of the two

identical circles inscribed inside the rectangle. Seethe diagram below.

Since the length is 3r and the width is 2r , the area

can be found by

3r × 2r = 6r 2

7. (D) 60 First, draw a diagram of the fan.

The length of the fan blade is the radius of a circle.

Find the distance traveled in one revolution by

finding the circumference of the circle.

C = 2r = 2(1.5 ft) = 3 ft

Since the fan spins 1,760 revolutions per minute, the

fan travels

(3 ft)(1,760) = 5,280 ft per minute

Since there are 5,280 feet per mile, the fan travels

miles per minute. There are 60 minutes in an hour,

the fan travels 60 miles per hour.

IX. Solid Figures

1. (D) 128 Start with the formula for volume of a

rectangular solid, V = lwh. Since the volume is 64

and the width is 12

,

64 =1

2lh

Multiply both sides of the equation by 2 to get

128 = lh

Since the area of the shaded face is equal to the

height of the figure times the length of the figure, the

area is 128.

2. (D) 216 Since the surface area of the cube is

given, start with the formula for surface area of a

cube, “Surface Area = 6 s2.” Using this formula,

solve for the length of the side, s.

6 s2 = 216

s2 = 36

s = 6

Now that the length of the side is known, find thevolume of the cube using the formula “Volume = s3.”

s3 = 63 = 216

(Note: The volume and surface area of a cube are

not typically the same number. This occurred here

only because the length of the side is 6. For any

r r

r

r

r

3r

2r

1.5 feet




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other given side length of a cube, the volume and

surface area will be different values.)

3. (A) 1 The maximum height of the cylinder

would be 10 in. Therefore, the 6 in. will form the

circumference of the base. Since the shape of the

base is a circle and the circumference is 6 in, use theformula “C = 2r ” to solve for the radius.

2r = 6

r =6 3

2 1

(Note: This problem did not ask for an exact value

of the radius. Approximate 3 for π.)

4. (D) 2l + 2w + 4h + 20 The ribbon crosses thelength of the box twice, the width twice, and the

height 4 times. The total length of the ribbon is then

twice the length, twice the width, and 4 times the

height of the box plus 20 inches for the bow.

Mathematically, this translates into

2l + 2w + 4h + 20

X. Advanced Volume – Solutions

1. (B) 38 cm3 The volume of a right circular cone is

given by the formula21

3V r h . Using this formula

and the information given by the diagram, compute

the volume. The height of the cone is 4 cm, so h = 4,

and the diameter is 6 cm, so the radius r = 3. Plug

these numbers into the formula to obtain:

V =1

3π (3)2(4) =

1

3π (36) = 12π.

Given the fact that π ≈ 3.14, 12π = 12(3.14) ≈ 38

cm3, or (B).

2. (B) 6 cm The question gives the formula for the

volume of a sphere and then gives the volume of one

such sphere. Given this information, it is possible to

compute the radius, since the question asks for the

diameter.

Set the formula equal to the given volume and solve

accordingly:

4

3πr

3 = 36π

4

3r 3 = 36

r 3 = 27

r = 3

The radius is 3 cm. However, note that the question

asks for the diameter , NOT the radius. To obtain the

diameter, simply double the radius. The diameter is

6 cm, or (B).

3. (C) 6 Since the container is filled to the top, the

first step to finding the weight of the slag is to

calculate the volume of the container. The container

is a half cylinder, so the formula for solving the

volume is V =1

2πr 2h. In the diagram, the height (10

ft.) and the diameter (8 ft.) are given, so r = 4 (half

of the diameter) and h = 10. Plug these numbers into

the formula:

V =

1

2 π (4)2

(10) =

1

2 π (160) = 80π ft3

Now that the volume has been obtained, calculate

the weight of the slag. The slag weighs 50 pounds

per ft3, so the weight of the slag in pounds is 50(80π)

= 4000π pounds.

Note, however, that the question asks for the weight

of the slag in tons. The conversion of 1 ton = 2000

pounds has been provided, so the weight of the slag

is 2π tons, which is about 6 tons, or (C).

XI. Coordinate Geometry

1. (D) (4, 3) The midpoint of any line segment is

equal to the average of the x-coordinates and the y-

coordinates. Find these averages.

( x, y) =4 4 2 8 8 6

, ,2 2 2 2

(4, 3)




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2. (B) l A slope is positive if it moves upward

from left to right. A slope is negative if it moves

downward from left to right. The greatest slope will

be one which is positive and has the sharpest incline

upwards. The least slope will be the one which is

negative and has the sharpest decline downwards.Line r has a negative slope. Line l has a positive

slope. Lines m and n are both positive, but neither

have a sharper incline than l . Line p has a slope of 0.

Therefore, line l has the greatest slope.

3. (D) 3 Using the formula for slope, “m =

2 1

2 1

y y

x x

”, solve for the missing coordinate.

m = 2 1

2 1

y y

x x

2

3 =

7

4 ( 2)

w

2

3 =

7

6

w

From here, cross multiply.

12 = 3(7 – w)

12 = 21 – 3w

3w = 9

w = 3

4. (C) 4 The longer side of the rectangle is equal

to the greater difference of the coordinates. The

difference between the x-coordinates is 5 – 1 = 4.

The difference between the y-coordinates is 4 – 2 =

2. Since 4 is greater than 2, the longer side is 4.

2 1

2 1

12 7 5

8 6 2

y y

x x

XII.Area of Many-SidedPolygons

SAMPLE PROBLEM: (A) 7 The sum of the

angles in an octagon is

(8 – 2)180º = 1080º

Since the figure is equiangular and there are 8

angles, every angle in the figure is equal to

1080

8

= 135º

Since every angle is equal to 135º , every angle can

be divided into a 45º and 90º angle. This allows the

octagon to be divided into five rectangles and four45º-45º-90º right triangles as shown in the figure.

Since each 45º-45º-90º right triangle has a

hypotenuse of 2 , the sides must have a length of 1.

Therefore, the area of each triangle is

A =1

2bh =

1

2(1)(1) =

1

2

Remember that the legs of every right triangle have

a length of 1. This means that the rectangles in the

figure are actually squares with a side length of 1.

Therefore, the area of each square is

A = s2 = (1)2 = 1

Since there are five squares and four triangles, the

entire figure has an area of

5(1) + 41

2

= 5 + 2 = 7

1

11

1

45º

45º

135º




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1. (A) 54 3 The sum of the angles in a hexagon

is

(6 – 2)180º = 720º

Since the figure is equiangular and there are 6

angles, every angle in the figure is equal to

720º

6 = 120º

Therefore, the figure can be divided into six

equilateral triangles as shown in the figure.

The base of each triangle is 6. The height of the

triangle can be determined by dividing the

equilateral triangle into two 30º-60º-90º triangles as

shown below.

The height of the 30º-60º-90º triangle must be 3

times the side opposite the 30º angle. Therefore the

height is 3 3 . Now, the area of each triangle can be determined.

A =1

2bh =

1

2(6) 3 3 = 9 3

Since there are six equilateral triangles, the total area

is

(6) 9 3 = 54 3

XIII. Trigonometry – Solutions

1. (A) 3 Recall that sin( x) =opposite

hypotenuse. Therefore,

sin( P ) =QR

QP . Since sin( P ) = 0.5 and QP = 6, this

means that

6

QR = 0.5

Multiply both sides by 6.

QR = 3

Solving this equation gives QR = 3, so (A) is the

answer.

2. (C) x Recall that cos( x) =adjacent

hypotenuse.

Therefore, in the diagram, cos = x

r . Multiply both

sides by r .

cos =r

r cos = x

The answer is (C).

3. (A) 2 Recall that sin( x) =opposite

hypotenuseand cos( x)

=adjacent

hypotenuse

. The question asks for the value of

sin( ) cos( )

cos( )

A B

B

. Solve for each piece. The sin( A)

=a

cusing the diagram. The cos( B) =

a

caccording to

the figure. Therefore:

120º

60º6

6

6

6

6

6

60º

6

3

30º

60º




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sin( ) cos( )

cos( )

A B

B

=

a a

c ca

c

Simplify by combininga

c

+a

c

.

a a

c ca

c

=

2a

ca

c

Remember that dividing by a fraction is equal to

multiplying by the reciprocal.

2a

ca

c

=2a c

c a

Simplify.

2a c

c a = 2

(A) is the answer.

XIV. Equation of a Circle –

Solutions

1. (B)2

( 5) ( 3) 16 x y The equation is a circle

is given by the formula ( x – x0)2 + ( y – y0)

2 = r 2,

where ( x0, y0) is the center of the circle and r is the

radius. The center of the circle is (5, -3) and the

radius is 4. Therefore, the equation of this circle is ( x

– 5)2 + ( y + 3)2 = (42) = 16, or (B).

2. (D) (8, -3) To find the coordinate of the other end

of the diameter, note that to move from the point A

(2, 1) to (5, -1), we moved 3 units right and 2 unitsdown. To find the other coordinate, move 3 units

right and 2 units down from the center point (5, -1)

to obtain (8, -3), or choice (D). This works because

the two points of the diameter are each the same

distance from the center of the circle and both lie on

the same line with the same slope.

Another way of solving this problem is to note that

the center of the circle is the midpoint of the two

points on the diameter. To find the midpoint,

average the x values and take the average of the y

values. This means that (2+x)/2 = 5 and (1+y)/2 = -

1. You can solve for ( x,y) to find that the other pointis at (8,-3).

3. (A) (0, 10) and (0, -10) The equation of a circle

centered at the origin is x2 + y2 = r 2. Given the

equation x2 + y2 = 100, 100 = r 2 and r = 10. The

points where this circle crosses the y-axis are given

by (0, r ) and (0, -r ). Since r = 10, the two points are

(0, 10) and (0, -10), so (A) is the correct answer.

MATH TRICKS

I. Whole minus Part

SAMPLE PROBLEM 1: (C) 4 –2

The shaded

region is equal to a square minus a semicircle. The

square has side lengths of 2. The semicircle has a

diameter of 2, so it has a radius of 1. The area of the

shaded region can now be found by

Area of square – Area of semi-circle

s2 –2

2

r

(2)2 –2(1)

2

4 –2

SAMPLE PROBLEM 2: (D) mn –2

4

m The

shaded region is equal to a rectangle minus two

semicircles. The width of the rectangle is m and the

length is n. The diameter of the semicircles is m, so

the radius is

2

m. Also, two semicircles are equivalent

to one whole circle. Therefore, the area of the

shaded region can be found by

Area of rectangle – Area of circle

lw – πr 2

mn – π2

2

m




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mn –2

4

m

1.(D) π The shaded area is equal to a semicircle

minus the non-shaded area in the middle. Since a

circle is 360º, the sector with 90º must be90

360

or

1

4 of the circle. Therefore, the area of the shaded

region must be equal to a semi circle minus one

fourth of the circle.

Area of semicircle – Area of1

4 of circle

2 2

2 4

r r

2 2

(2) (2)2 4

4 4

2 4

2π – π = π

2. (A) 4 – π The shaded area is equal to the area of

the square minus a section of the circle. Since all

angles in a square are 90º, the section of the circle

must have an angle of 90º. A circle is 360º, so this

section of the circle must be90

360

or

1

4 of the circle.

Therefore, the shaded region is the area of the square

minus1

4 of the circle. Also, notice that the radius

of the circle is equal to the length of a side of the

square.

Area of the square – Area of1

4 of circle

s2 –2

4

r

(2)2 –

2

(2)4

4 –4

4

= 4 – π

3. (B) 9π – 18 Since the marked angle in the

figure is 90º, this section of the circle must be90

360

or1

4 of the circle. The shaded area is equal to one

fourth of the circle minus the area the right triangle

ABC . Notice that the base and height of the triangle

ABC are both equal to the radius of the circle.

Area of1

4 of circle – Area of triangle

2

4

r –

1

2bh

2(6)

4

–

1

2(6)(6)

36

4

– 18 = 9π – 18

II. Guesstimating

SAMPLE PROBLEM 1: (A) 50(π – 2) The

shaded area is equal to the square ABCD minus the

two white “curved triangles.” First, find the area of

one of the “curved triangles.” One “curved triangle”

equals the area of the square minus the area of1

4 a

circle with radius 10.

Area of square – Area of1

4

a circle

s2 –2

4

r

(10)2 –2(10)

4

100 –100

4

100 – 25π

Again, the area of the shaded region is equal to the

area of the square minus the area of two “curved

triangles.”

Area of the square – Area of two “curved triangles”

s2 – 2(100 – 25π)

102 – 2(100 – 25π)

100 – 200 + 50π 50π – 100

50(π – 2)




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SAMPLE PROBLEM 2: (A) 6 3 – 6 In the

triangle ABD, if angle A is 30º and angle B is 90º,

angle D must equal 60º because the sum of the

angles of a triangle equals 180º. Therefore, triangle

ABD is a 30º-60º-90º right triangle. If segment AD is

12, then segment BD is 6 and segment AB is 6 3 .

Since triangle ABC is an isosceles triangle, segment

BC is equal to segment AB which is 6 3 . Notice

that segment DC is equal to segment BC minussegment BD.

DC = BC – BD

DC = 6 3 – 6

1. (C) 120 The angle above the line is equal to 90º

minus 30º, or 60º. Remember that the degree

measure of a straight line is 180º. Therefore,

xº + 60º = 180º xº = 120º

2. (C) 90 Remember that when lines intersect, two

pairs of equal angles are formed. The equal angels

are shown below.

Remember that a triangle has 180º. Therefore,

xº + 45º + 45º = 180º

xº + 90º = 180º

xº = 90º

3. (B) 45 This problem is designed to test the

ability to guesstimate angles. (Note: This type of

problem will not appear on an actual exam. It is only

an exercise to help practice guesstimating.) This

triangle appears to have a 90º angle and two equal

angles, of which x is one of them. It therefore

appears to be a 45º-45º-90º triangle and the best

answer is 45º.

4. (C) 40 Since MN is parallel to OP , angle ACP

is equal to angle xº, which is 100º. Since the triangle

is equilateral, angle ACB is 60º. Notice that angle

ACP is equal to angle ACB plus angle yº.

ACP = ACB + yº

100º = 60º + yº40º = yº

5. (D) 360 Label the interior angles of the triangle

as aº, bº, and cº as shown below.

Since a straight line is 180º, notice that each pair of

angles ( xº and bº, yº and aº, z º and cº) adds up to

180º. Therefore, an equation can be formed.

aº + bº + cº + xº + yº + z º = 180º + 180º + 180º

aº + bº + cº + xº + yº + z º = 540º

Since a triangle is 180º, then aº + bº + cº = 180º.

Now, replace aº + bº + cº with 180º.

180º + xº + yº + z º = 540º

xº + yº + z º = 360º

Guesstimating – PracticeProblems

(Note: Problems 1 through 9 are designed to test the

ability to guesstimate. Problems of this sort will not

appear on an actual exam, but the ability to

guesstimate may help with certain problems on an

exam.)

1. (D) 75 The angle here is less than 90º butgreater than 45º. While 60º is an acceptable guess,

the angle is closer to 75º.




13/13

2. (A) 120 The angle here is greater than 90º.

While 95º is an acceptable guess, the angle is closer

to 120º.

3. (E) 90 The angle here appears to be exactly 90º.

4. (A) 15 The angle made up of both x’s looks likeabout 30º together, so each x is likely 15º.

5. (A) 90 The angle here appears to be 90º.

6. (C) 175 The angle here is a little less than half

of a circle. Since a circle is 360º, half of a circle is

180º. Therefore, the angle here is a little less than

180º , making the best guess 175º

7. (B) 120 The angle here appears to be1

3 of a

circle. Since a circle is 360º,1

3 of a circle is 120º.

Therefore, the angle here is likely 120º.

8. (C) 45 The angle here is less than 45º. While

15º is an acceptable guess, the angle is closer to 30º.

9. (D) 720 The triangle appears to be equilateral,

meaning each angle is 60º. Each of the indicated

angles would then be 120º because the degree

measure of a line is 180º, so each angle would be180º – 60º = 120º. The sum would then be 6 × 120º= 720º.


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